3
$\begingroup$

The GAIA spacecraft rotates at 1 degree per minute about its axis, scanning two telescopes looking 106.5 degrees apart but imaging on the same focal plane around a circle every 6 hours.

The rotation axis is kept at 45 degrees with respect to the direction of the Sun, but it is said that it precesses around an axis pointed at the Sun once every 63 days (4 degrees per day).

Note that GAIA is in a Lissajous orbit around Sun-Earth L2 with a period of 180 days, and of course L2 orbits the Sun every 365 days, so the direction that points towards the sun also moves; this so-called "precession" is not about an inertial axis! I'm totally confused by all of this.

Figures 6 and 7 in The Gaia mission (also here) as well as all the math there give some insight into the motion, but not necessarily the reason. There is some discussion of the spacecraft in this paper as well.

GAIA's astrometric performance is at the micro-arcsecond level

I don't think this is done by nudging with thrusters, there must be some elegant and super-smooth (and probably passive) way that this was implemented.

Question: What causes GAIA rotational axis to precess the way it does? How exactly is this accomplished in such a super-smooth way?


enter image description here

Source


$\endgroup$
2
  • $\begingroup$ $\mathbf I^{-1} ( \mathbf I \omega \times \omega)$ $\endgroup$ Apr 8 at 10:21
  • $\begingroup$ @DavidHammen the info I've quoted says GAIA's spin axis precesses around an axis pointing at the Sun. I've explained that one of the many challenges I'm having with this is that that axis is not inertial; it moves in a "loopy" way as GAIA moves around L2 and L2 moves around the Sun. I'll admit that I've always been rigid-body mechanics-challenged, this one is particularly hard for me. $\endgroup$
    – uhoh
    Apr 8 at 10:50
3
$\begingroup$

From the perspective of an inertial frame of reference, the change in an object's intrinsic angular momentum $\vec L$ is related to the external torque $\vec \tau$ on that object via the rotational equivalent of Newton's second law, $\dot{\vec L} = \vec \tau$. For a rigid body, $\vec L=\mathbf I\,\vec\omega$. Differentiating this with respect to time results in $\dot{\vec L} = \dot{\mathbf I}\,\vec\omega + \mathbf I\,\dot{\vec\omega}$. This is a bit ugly.

This is ugly because unlike mass (which is presumably constant), the inertia tensor $\mathbf I$ is not constant from the perspective of an inertial frame of reference. The inertia tensor of a constant mass rigid body is however constant from the perspective of a frame that rotates with the rigid body. Assuming a constant inertia tensor from the perspective of a frame that rotates with the object and applying the transport theorem ($\dot{\vec X}_\text{inertial} = \dot{\vec X}_\text{rotating} + \vec \omega \times \vec X$), the time derivative of angular momentum becomes $$\dot{\vec L}_{\text{rotating}} = \mathbf I\,\dot{\vec \omega} = \vec \tau_\text{ext} - \vec \omega \times \vec L = \vec \tau_\text{ext} - \vec \omega \times (\mathbf I\,\vec\omega) = \vec \tau_\text{ext} + (\mathbf I\,\vec\omega)\times \vec \omega$$ This is useful because $\dot{\vec\omega}$ is the same vector (sans a rotation) in the inertial frame and the rotating frame, once again thanks to the transport theorem. Applying the transport theorem to angular velocity results in $\dot{\vec \omega}_\text{inertial} = \dot{\vec \omega}_\text{rotating} + \vec \omega \times \vec \omega$. Since $\vec \omega \times \vec \omega$ is identically zero, this simplifies to $\dot{\vec \omega}_\text{inertial} = \dot{\vec \omega}_\text{rotating}$.

This remains confusing because, as Herbert Goldstein wrote in the very widely used text on classical mechanics, "hence the jabberwockian sounding statement: the polhode rolls without slipping on the herpolhode lying in the invariable plane." The behavior for an arbitrary body can be rather complex. Fortunately, the behavior becomes much simpler for a torque-free object with axial symmetry that rotates about an axis that is close to but not the same as the axis of symmetry. The axis of rotation precesses, with $\mathbf I^{-1}((\mathbf I\,\vec \omega)\times \vec \omega)$ describing the precession rate.

Spacecraft designers have a limited set of choices regarding the inertial torque $(\mathbf I\,\vec \omega)\times \vec\omega$ term:

  • They can fight the inertial torque with thrusters, but this typically is not the best idea.
  • They can make the spacecraft rotate as little as possible with regard to inertial space, thereby making the inertial torque be rather small.
  • They can make the spacecraft rotate close to one of the spacecraft's principal axes, once again making the inertial torque be rather small.
  • They can try to balance the external torques and inertial torque so as to keep the total torque on the vehicle very small.
  • They can take advantage of the inertial torque so as to make the spacecraft precess at a desired rate, with limited corrections from the attitude control system.

The designers of Gaia and its predecessor (Hipparcos) chose the final option.

$\endgroup$
1
  • $\begingroup$ Jabberwockian indeed! So when I am done reading and understanding this I'll potentially be able to calculate (though perhaps not understand) how Gaia's cleverly devised moment of inertia tensor causes it to precess 4 °/day about an axis that itself rotates about 1 °/day? An this is brought to you by the makers of the transport theorem? $\endgroup$
    – uhoh
    Apr 8 at 12:40
2
$\begingroup$

Precession of spinning objects

A spinning object can naturally precess, as in the case of GAIA satellite, only if its principal moments of inertia normal to the spin axis are equal to each other, and different from the principal moment of inertia parallel with the spin axis. That is, if the satellite spins around its Z axis (angular velocity vector in satellite's body reference frame $\vec{\omega} = [0, 0, \omega_z]^T$ is parallel with principal moment of inertia $I_z$), then it must be that $I_x = I_y \neq I_z$. This is necessary to keep the angular velocity vector locked in the body reference frame. The equation from David Hammen's answer can be broken down into three separate equation components for each of the three axes:

$\dot{\omega}_x = \frac{\tau_{ext_x} - \omega_y\omega_z(I_z - I_y)}{I_x}$

$\dot{\omega}_y = \frac{\tau_{ext_y} - \omega_x\omega_z(I_x - I_z)}{I_y}$

$\dot{\omega}_z = \frac{\tau_{ext_z} - \omega_x\omega_y(I_y - I_x)}{I_z}$

If it would be the case that $I_x = I_y = I_z$, then the inertial torque terms would disappear and the equations would become:

$\dot{\omega}_x = \frac{\tau_{ext_x}}{I_x}$

$\dot{\omega}_y = \frac{\tau_{ext_y}}{I_y}$

$\dot{\omega}_z = \frac{\tau_{ext_z}}{I_z}$

In this case the angular velocity vector components can be freely and independently changed by applying torques. A torque to one component will not change the angular velocity of another component. Also, notice that any torque application will necessarily change the angular velocity vector in the body reference frame. It is still possible to force the spinning object of this case to precess, which we will return to after we examine the $I_x = I_y \neq I_z$ case, where the precession happens naturally. There the inertial torque disappears only for the Z component, while for components X and Y we can introduce a constant $k = \omega_z(I_z - I_y) = -\omega_z(I_x-I_z)$. Notice the minus sign, as the $I_z$ position is changed. This gives following equations:

$\dot{\omega}_x = \frac{\tau_{ext_x} - \omega_y k}{I_x}$

$\dot{\omega}_y = \frac{\tau_{ext_y} + \omega_x k}{I_y}$

$\dot{\omega}_z = \frac{\tau_{ext_z}}{I_z}$

Now, if there is no spin around the Z axis ($\omega_z = 0$), then it is the same as in the previous case, but as soon as there is some spin ($\omega_z \neq 0$), the X and Y components become coupled. The greater the spin, the stronger the coupling. Because the signs before the inertial torque terms are different and the constant is same, the components will act against each other. Start with $\omega_x = 0$, $\omega_y = 0$ and $\omega_z$ is some large positive value, and introduce a small torque in X component. This will produce positive $\omega_x$, which then produces positive $\omega_y$; a positive $\omega_y$ will act to decrease $\omega_x$. With sufficiently high spin rate and large k, the X component torque is overpowered and $\omega_x$ quickly goes to negative. Now, a negative $\omega_x$ will act to decrease $\omega_y$. With a large k, $\omega_y$ is soon negative, and a negative $\omega_y$ acts to increase $\omega_x$. This is a little bit sketchy explanation, the reality involves some infinitesimals and makes the $\omega_x = 0$ and $\omega_y = 0$ a stable state, but hopefully should provide a bit of intuition. This stability is only regarding the body reference frame, one can still change the angular velocity vector freely in all three components in inertial reference frame. In fact, this is what drives precession, the torque applied and work it generates does not disappear in the thin air, it changes the orientation of the velocity vector while the axis of rotation of the body remains the same.

For the curious, the same applies whether $I_x = I_y < I_z$ or $I_x = I_y > I_z$ holds. Only the signs in X and Y components will switch places.

Back to the $I_x = I_y = I_z$ case, to make it precess, the control torque would have to imitate inertial torque terms with strength that overpowers the regular torque terms.

And for the completeness sake, the case $I_x \neq I_y \neq I_z$ does not have a natural precession as the inertial torque terms will be different for all three components and the Z component would not be decoupled. Still a control law could be made to force it precess.

Another important thing to consider is how the precession can be controlled. Or how is the angular velocity vector changed in the inertial reference frame? I do not believe that any amount of equations will be better than having a visual aid for understanding this. Luckily, there is a good number of videos online trying to explain. Here is one that have I randomly chose and found it to be pretty good

. One thing I will draw focus to in the video is that in the precessing case, the torque is changing and precessing together with the angular velocity vector. The act of precessing the object also moves where the force will act, even though the force is always in the same direction. This in turn generates the torque normal to both the force and the angular velocity vector. Here the reality, again, needs infinitesimals to produce a circular motion of the precession. As long as the torque is constantly normal to the angular velocity vector, it will precess, without changing the magnitude. It does not need to be circular, but here that is accomplished with the constant force. So far, I have been avoiding to talk about angular momentum, as for the case that interests us here it is always parallel with the velocity vector and only its scalar multiple. Still, it is important to keep in mind that the torque applied for precessing is limited so that it does not significantly modify angular momentum at any time. This is related to the inertial torque overpowering externally applied torque from the components equations. Too strong applied torque would be able to force the angular velocity vector out of its stable point. This could also be the case of relatively slow spin around the Z body axis and then the torque normal to the angular velocity vector would not be sufficient to create precession, but the torque would have to have a component along the angular velocity vector too.

GAIA satellite case

I was quite intrigued when I first saw that picture in your question. The Sun and the photon pressure from it would be perfect to naturally generate precession without the need of active attitude control (in ideal case that is, in reality it would of course still need to apply at least small attitude corrections especially with the 1 arcsec level of precision). But the excitement quickly went down realizing that the torque from the solar pressure would be very small.

GAIA, by looking at its pictures, probably fits the case of $I_x = I_y < I_z$. In https://www.researchgate.net/publication/258736355_Attitude_reconstruction_for_the_Gaia_spacecraft the $I_z = 4500 kgm^2$ is provided.

The angular velocity of $\omega_z = 1 deg/min$ sounds quite slow, but it is 360 times larger than the precesion rate of $\omega_p = 4 deg/day$. This is somewhat optimistic that we can have natural precession. The angular momentum is $L = I_z \omega_z = 1.018 kg m^2 s^{-1}$.

A formula connecting angular velocity, precession rate and a torque when $\omega_z >> \omega_p$ (a derivation can be found here https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/11-3-precession-of-a-gyroscope/) is:

$\omega_p = \frac{\tau}{Lsin\theta}$

where $\theta$ for GAIA satellite is $45 deg$. Thus, the required torque is $\tau = 1.999 m N m$. One second of this torque applied to the satellite would change its angular momentum no more than 0.2%. I think this means that the natural precession can be achieved with a torque normal to both the angular velocity vector and the Sun direction vector.

From https://www.researchgate.net/publication/257561746_Dynamical_attitude_model_for_Gaia we can learn that the solar pressure torque (which is also the greatest external torque source, so we do not have to care of other torque sources for this analysis) is $\tau_{sp} = 134 u N m$, almost 15 times weaker than necessary. This means that the GAIA satellite must constantly and actively provide control torque.

Also, the article talks about canceling the solar pressure torque, so it is probably not even the case that the satellite is designed to leverage the external source, but has to rely on its micropropulsion system to first cancel the external torque and then to control the attitude. It is also very interesting that the effects from micrometeroid impacts are quite significant, relative to mission's attitude error tolerances, for which the attitude system needs to constantly do corrections. I had opportunity, several days ago, to attend a presentation where in-flight data of micrometeroids impacts were shown. In the plots it was also noticeable a fine constant wobbling of attitude, before impacts, that comes from the micropropulsion corrections. If I understood that correctly, it is another confirmation that the attitude system is constantly active.

$\endgroup$
1
  • $\begingroup$ Thanks for your answer! It will take some time for me to go through it. There is some discussion of solar pressure torques on Gaia in this comment and which also links here. $\endgroup$
    – uhoh
    Apr 12 at 21:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.